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Statement-1: The number of zeros at the ...

Statement-1: The number of zeros at the end of 100! Is, 24.
Statement-2: The exponent of prine p in n!, is
`[(n)/(p)]+[(n)/(p^(2))]+.......+[(n)/(p^(r))]`
Where r is a natural number such that `P^(r)lenltP^(r+1)`.

A

Statement-1 is True, Statement-2 is true, Statement-2 is a correct explanation for Statement-1.

B

Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.

C

Statement-1 is True, Statement-2 is False.

D

Statement-1 is False, Statement-2 is True.

Text Solution

AI Generated Solution

To determine the number of zeros at the end of \(100!\), we need to find the number of times \(10\) is a factor in \(100!\). Since \(10 = 2 \times 5\), we need to find the minimum of the number of factors of \(2\) and \(5\) in \(100!\). ### Step-by-Step Solution: 1. **Finding the number of factors of 2 in \(100!\)**: - We use the formula for the highest power of a prime \(p\) in \(n!\): \[ \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor ...
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